frac{{cos 12^circ - sin 12^circ }}{{cos 12^ci | vedclass

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Question

(C) Given expression: $\frac{{\cos 12^\circ - \sin 12^\circ }}{{\cos 12^\circ + \sin 12^\circ }} + \frac{{\sin 147^\circ }}{{\cos 147^\circ }}$Divide

Vedclass · Accepted Answer

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Vedclass · Answer

(C) Given expression: $\frac{{\cos 12^\circ - \sin 12^\circ }}{{\cos 12^\circ + \sin 12^\circ }} + \frac{{\sin 147^\circ }}{{\cos 147^\circ }}$Divide the numerator and denominator of the first term by $\cos 12^\circ$:$= \frac{{1 - 	an 12^\circ }}{{1 + 	an 12^\circ }} + 	an 147^\circ $Using the formula $	an(45^\circ - A) = \frac{{1 - 	an A}}{{1 + 	an A}}$,the first term becomes $	an(45^\circ - 12^\circ) = 	an 33^\circ$.For the second term,$	an 147^\circ = 	an(180^\circ - 33^\circ) = -	an 33^\circ$.Substituting these values back into the expression:$= 	an 33^\circ + (-	an 33^\circ) = 0$.

$\frac{{\cos 12^\circ - \sin 12^\circ }}{{\cos 12^\circ + \sin 12^\circ }} + \frac{{\sin 147^\circ }}{{\cos 147^\circ }} = $

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